how is that expression is generated

Question

I have no idea that how is red arrowed mammoth term is generated from the yellow arrowed term

Please explain

Answer 1

The author starts by substituting $t$ for $e^x - 1$, hence $$ \sin (e^x - 1) = \sin t $$ Now the power series of the sine function is plugged in (yellow arrowed line on the right) $$ \sin (e^x - 1) = \sin t = t -\frac{t^3}{3!} + \frac{t^5}{5!} \mp \cdots $$ Now we resubstitute $e^x -1$ for $t$, at the same time we use the power series expansion of the exponential function $$ e^x = 1 + x + \frac{x^2}2 + \frac{x^3}{6} + \cdots $$ or $$ e^x - 1 = x + \frac{x^2}2 + \frac{x^3}{6} + \cdots $$ giving $$ \sin (e^x - 1) = \left(x + \frac{x^2}2 + \frac{x^3}{6} + \cdots\right) - \frac 16\left(x + \frac{x^2}2 + \frac{x^3}{6} + \cdots\right)^3 + \frac 1{120}\left(x + \frac{x^2}2 + \frac{x^3}{6} + \cdots\right)^5 \mp \cdots $$ which is the red arrowed expression.

Answer 2

This is just a substitution. You have from Taylor Series $$ \sin(t) = t- \dfrac{t^3}{3!} + \dfrac{t^5}{5!}- \dfrac{t^7}{7!} + \cdots= t- \dfrac{t^3}{6} + \dfrac{t^5}{120}- \dfrac{t^7}{5040} + \cdots $$ But you know also the Taylor Series for $e^x-1$: $$ e^x-1 = x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots $$ which the book says to call this $t$. But then we plop this Taylor Series where we see $t$ in the Taylor Series for $\sin t$: $$ \begin{split} \sin(e^x-1)&= \sin(t) \\ &= t- \dfrac{t^3}{3!} + \dfrac{t^5}{5!}- \dfrac{t^7}{7!} + \cdots \\ &= t- \dfrac{t^3}{6} + \dfrac{t^5}{120}- \dfrac{t^7}{5040} + \cdots \\ &=\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)- \dfrac{\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)^3}{6} + \dfrac{\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)^5}{120}- \dfrac{\left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\right)^7}{5040} + \cdots \end{split} $$

Answer 3

In the yellow term, you let $t = e^x-1$, and then consider the Taylor expansion of $\sin t$, i.e. $$\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} -\cdots$$ Now, instead of each $t$ on the right-hand side, you want to substitute $t=e^x-1$, but instead of just plugging in $e^x-1$ instead of $t$, you plug in the Taylor expansion of $e^x-1$, which is $$e^x-1 = x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$$

So each $t$ is replaced by this expression, which gives you the mammoth term.

Answer 4

Maclaurin series is a Taylor series expansion of a function about $0$,
Any function f(x) can be written as:

$$f(x)=f(0)+f^{'}(0)x+ \frac{f^{''}(0)}{2!}x^2+.....$$ (where $f^{'''n}$ denotes $n^{th}$ derivative of f(x))

Take $$f(x)=\sin x$$ Take derivatives and put in above equation.

You will get $$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$

For your question, $$e^x-1=t$$
So, just put $$e^x-1$$ in place of x in above formula, you will get to the desired result.